Question

An amount of water of mass $20$g at $0^o$C is mixed with $40$g of water at ${10}^o$C, then the final temperature of mixture is :

Answer 1

We know,

$T^{\prime }=\frac{m_1{T}_1+m_2{T}_2}{m_1+m_2}=\frac{20\times 0+40\times 10}{20+40}=6.6°C$

is the required final temperature.